Lesson 17 ~25 min Unit 2 · Non-Linear +85 XP

Applications of Non-Linear Relationships

Parabolas for projectiles, hyperbolas for inverse proportion, exponentials for growth, circles for circular paths. Read graphs in context and answer real-world questions.

Today's hook: A ball is thrown up and lands $4$ seconds later. Its height $h = -5t^2 + 20t$. When is it highest? How high does it get? What family is $h(t)$?

0/5QUESTS

Printable Worksheets

Print or save as PDF — or build a custom worksheet from any module's questions.

Build Apply Master Build custom

Think First

warm-up

Non-linear graphs are everywhere in real life: projectile motion (a kicked ball), inverse proportion (more workers, less time), bacteria growth, racetracks. For each scenario, name the family you'd expect: (a) a stone tossed in the air; (b) doubling rabbits each month; (c) trip time vs speed for a fixed distance; (d) the path of a runner on a circular track.

Record your answer in your workbook.

The Big Idea

+5 XP

Each family describes a different kind of real-world relationship. Recognising the family from the context lets you predict the shape, read key information off the graph and answer questions in real units.

Projectile — height vs time is a parabola: max height = vertex, landing = $x$-intercepts. Inverse proportion — $y = \dfrac{k}{x}$: doubling one halves the other. Population growth/decay — exponential: each step multiplies. Circular path — circle equation: every point is the same distance from the centre.

Family fixes shape; shape tells the story.

Real units

Axes have units: m, s, $, people — always.

Read features in context

Vertex = max height. $x$-int = landing. Asymptote = limit.

Check domain

Time can't be negative; populations can't be fractional people.

What You'll Master

objectives

Know

Which non-linear family models which real-world scenario
How to read max/min, intercepts, asymptotes in context
How to apply domain restrictions in word problems

Understand

Why height-vs-time under gravity is a parabola
Why fixed distance + variable speed gives inverse proportion
Why "doubles each step" is exponential growth

Can Do

Identify the family from a worded scenario
Translate a feature (vertex, $x$-int, asymptote) into real-world meaning
Answer "at what time", "what value", "how long until" questions from a graph

Words You Need

vocabulary

ProjectileAnything thrown/launched under gravity; height vs time is a parabola.

Inverse proportion$y = \dfrac{k}{x}$ — doubling $x$ halves $y$; product $xy = k$ stays constant.

Exponential growthQuantity multiplies by a constant factor each step (e.g. doubles every hour).

Exponential decaySame idea, but factor between $0$ and $1$ (e.g. halves each step).

Domain restrictionReal-world constraint (e.g. time $\ge 0$, population whole numbers).

Maximum heightThe $y$-coordinate of the vertex when a parabola opens down.

Spot the Trap

heads-up

Wrong: Reporting a max height of "$20$" without units.

Right: "Max height is $20$ metres at time $t = 2$ seconds". Always include units.

Wrong: Using both $x$-intercepts $t = -1$ and $t = 5$ as landing times.

Right: Time can't be negative. Reject $t = -1$; the ball lands at $t = 5$ seconds.

Family-to-Scenario Map

+5 XP

The four families each model very different real-world relationships:

Parabola: projectile motion ($h = -\tfrac{1}{2}g t^2 + v_0 t + h_0$), bridge arches, headlight reflectors, profit functions.
Hyperbola ($y = \dfrac{k}{x}$): fixed-distance trips (time vs speed), constant volume (pressure vs volume), people splitting a fixed cost.
Exponential ($y = a \cdot b^x$): compound interest, population growth, bacteria, radioactive decay (when $b < 1$).
Circle ($x^2 + y^2 = r^2$): running tracks, satellite orbits (idealised), Ferris wheels.

Read the scenario, name the family, predict the graph shape.

"Thrown" / "kicked"

Parabola: vertex = peak.

"Fixed total / amount"

Hyperbola: more of one $\Rightarrow$ less of the other.

"Doubles" / "halves"

Exponential: factor each step.

Reading Features in Context

+5 XP

Once you've identified the family, translate every graph feature into the scenario's language:

Parabola (projectile): $y$-intercept = launch height. $x$-intercepts = landing times (reject negatives). Vertex = (time of max, max height).
Hyperbola (fixed distance): $x$-axis = speed, $y$-axis = time. As speed $\to \infty$, time $\to 0$. As speed $\to 0$, time $\to \infty$.
Exponential (growth): $y$-intercept = starting amount. Asymptote = lower limit (often $0$ for decay). Doubling time = time to multiply by $2$.

Feature $\to$ context phrase. Always state with units.

Vertex = max/min

Max height, max profit, min cost — depends on context.

Reject impossible

Negative time, negative people: discard.

Always include units

Marks lost for naked numbers.

Watch Me Solve It · 3 examples

Watch Me Solve It · Projectile

+15 XP per step

PROBLEM

A ball is thrown up with height $h = -5t^2 + 20t$ (m, s). (a) When is the ball at maximum height? (b) What is the max height? (c) When does it land?

1
Identify family and axis
Parabola, opens down ($a = -5 < 0$). Axis of symmetry: midway between $x$-intercepts. Set $h = 0$: $-5t^2 + 20t = 0 \Rightarrow -5t(t - 4) = 0 \Rightarrow t = 0$ or $t = 4$.
2
Time of max height
Axis is midway: $t = \dfrac{0 + 4}{2} = 2$ s. So max occurs at $t = 2$ seconds.
3
Max height and landing
Sub $t = 2$: $h = -5(4) + 20(2) = -20 + 40 = 20$ m. Max height $= 20$ m. Lands when $h = 0$ again: $t = 4$ seconds (reject $t = 0$, that's launch).
Vertex tells you "best", $x$-intercepts tell you "start and end".

Answer(a) $t = 2$ s, (b) $20$ m, (c) $t = 4$ s.

Watch Me Solve It · Inverse proportion

+15 XP per step

PROBLEM

A driver travels $120$ km. Time $T$ (hours) vs average speed $v$ (km/h). (a) Write the relation. (b) Sketch the shape. (c) How long does it take at $60$ km/h? At $30$ km/h?

1
Build the relation
Distance = speed $\times$ time, so $120 = v \cdot T \Rightarrow T = \dfrac{120}{v}$. Inverse proportion — hyperbola in the $(v, T)$ plane.
2
Shape and domain
Only $v > 0$ makes sense (speeds are positive). One branch in Quadrant 1. Asymptotes $v = 0$ (vertical) and $T = 0$ (horizontal).
3
Substitute
At $v = 60$: $T = \dfrac{120}{60} = 2$ hours. At $v = 30$: $T = \dfrac{120}{30} = 4$ hours.
Halving speed doubles time — classic inverse proportion.

Answer$T = \dfrac{120}{v}$. At $60$ km/h: $2$ h. At $30$ km/h: $4$ h.

Watch Me Solve It · Population growth

+15 XP per step

PROBLEM

A bacteria colony starts at $100$ cells and DOUBLES every hour. Let $N(t) = 100 \cdot 2^t$, $t$ in hours. (a) How many cells at $t = 0$, $1$, $3$? (b) When does the population reach $1600$ cells?

1
Identify family
Exponential growth, base $2$, starting value $100$. $N(0) = 100 \cdot 2^0 = 100$. $N(1) = 200$. $N(3) = 100 \cdot 8 = 800$.
2
Set up for $1600$
$100 \cdot 2^t = 1600 \Rightarrow 2^t = 16$.
3
Solve by inspection
$2^4 = 16$, so $t = 4$ hours.
Inspection works whenever the target is a small power of the base.

Answer(a) $100, 200, 800$. (b) $t = 4$ hours.

Common Pitfalls

heads-up

Missing units

Writing "max height = $20$" without "metres" loses marks.

Fix: every numeric answer needs units in word problems: m, s, m/s, $, people.

Accepting impossible solutions

Reporting both $t = -2$ and $t = 5$ as landing times.

Fix: REJECT negative times, negative populations, fractional people. State why.

Wrong family for the context

Using a parabola for population growth or an exponential for a thrown ball.

Fix: ask — is it multiplying (exp), is gravity pulling (parabola), is the product fixed (hyperbola)?

Copy Into Your Books

Projectile

$h = -\tfrac{1}{2}g t^2 + v_0 t + h_0$
Parabola (opens down)
Vertex = max height
$x$-int = launch / land

Inverse proportion

$xy = k$, so $y = \dfrac{k}{x}$
Hyperbola, Q1 only (positive only)
Asymptotes both axes
Double one $\to$ halve other

Exponential growth

$N = N_0 \cdot b^t$, $b > 1$
$N_0$ = starting amount
Asymptote: $N = 0$ (decay)
Doubling time depends on $b$

Reading + units

Always state units
Reject negative time/people
Vertex $\to$ max/min
$y$-int $\to$ initial value

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Real-World

4 problems

Four quick problems on context interpretation.

1 A stone is thrown so $h = -5t^2 + 30t$. Find the time it lands ($h = 0$, $t > 0$).
$-5t(t - 6) = 0 \Rightarrow t = 0$ (launch) or $t = 6$ (land).$t = 6$ seconds
2 Six workers do a job in $10$ days. Assuming inverse proportion, how long for $3$ workers?
$k = 6 \times 10 = 60$ worker-days. $T = \dfrac{60}{3}$.$20$ days
3 A colony of $50$ doubles every hour: $N = 50 \cdot 2^t$. Find $N$ at $t = 4$.
$50 \cdot 2^4 = 50 \cdot 16$.$N = 800$ cells
4 A circular track has equation $x^2 + y^2 = 100$ (m). State the radius and a point on the track due north of the centre.
$r = \sqrt{100} = 10$. Due north of centre $(0,0)$ is $(0, 10)$.Radius $10$ m; point $(0, 10)$

Complete in your workbook.

Quick Check · 5 questions

Which family models the height of a thrown ball over time?

+10 XP

For a fixed-distance trip, time vs speed is a:

+10 XP

For $h = -5t^2 + 20t$, the height at $t = 1$ s is:

+10 XP

A culture starts at $200$ cells, doubling every hour. After $3$ hours, $N = $:

+10 XP

A projectile equation factorises so $h = 0$ when $t = -1$ or $t = 5$. The landing time is:

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. A diver's height (in metres) above water $t$ seconds after a $10$ m platform jump is modelled by $h = -5t^2 + 5t + 10$. (a) Find the diver's launch height. (b) Find when $h = 0$ (entry to water). (c) State the maximum height.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. Six friends share a $\$300$ accommodation cost. Each friend's cost $C$ (dollars) is inversely proportional to the number of friends $n$. (a) Write $C$ in terms of $n$. (b) Compute $C$ for $n = 3$, $n = 6$, $n = 10$. (c) What happens to $C$ as $n$ gets very large?

Answer in your workbook.

ReasonHard3 MARKS

Q8. A bank account starts at $\$500$ and grows by a factor of $1.1$ each year: $A = 500 \cdot 1.1^t$. (a) Compute $A$ after $1$ year and $2$ years. (b) Explain why this is an exponential growth model. (c) The model gives $A \approx 805.26$ at $t = 5$. What does the value at $t = 0$ tell you in the real-world?

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — parabola (gravity gives a quadratic in $t$).

2. B — hyperbola ($T \cdot v = d$ constant).

3. A — $h(1) = -5 + 20 = 15$ m.

4. D — $200 \cdot 2^3 = 1600$.

5. B — reject $t = -1$; land at $t = 5$ s.

Show Your Working Model Answers

Q6 (3 marks): (a) $h(0) = -5(0) + 5(0) + 10 = 10$ m [1]. (b) $-5t^2 + 5t + 10 = 0 \Rightarrow -5(t^2 - t - 2) = 0 \Rightarrow t^2 - t - 2 = 0 \Rightarrow (t - 2)(t + 1) = 0$, so $t = 2$ s (reject $t = -1$) [1]. (c) Axis: midway between $t = -1$ and $t = 2$ is $t = 0.5$. $h(0.5) = -5(0.25) + 5(0.5) + 10 = -1.25 + 2.5 + 10 = 11.25$ m [1].

Q7 (3 marks): (a) $C = \dfrac{300}{n}$ [1]. (b) $C(3) = \$100$, $C(6) = \$50$, $C(10) = \$30$ [1]. (c) As $n \to \infty$, $C \to 0$ — per-person cost approaches zero (horizontal asymptote $C = 0$) [1].

Q8 (3 marks): (a) $A(1) = 500 \cdot 1.1 = \$550$, $A(2) = 500 \cdot 1.21 = \$605$ [1]. (b) Each year the amount is MULTIPLIED by the fixed factor $1.1$ (a $10\%$ increase). Repeated multiplication is the defining feature of an exponential [1]. (c) $A(0) = 500$ is the initial deposit — the balance at the moment the account opened, before any interest [1].

Stretch Challenge · +25 XP, +10 coins

Match Real-World to Family

For each scenario, (i) name the family, (ii) write an equation in $x$, (iii) state what the $y$-intercept (or asymptote) means in context.
(a) A water tank of $1200$ L empties at $20$ L/min — volume vs time.
(b) A bridge arch is parabolic, $20$ m wide at the base, $8$ m tall at the centre.
(c) A radioactive sample halves every $5$ years from an initial $80$ g.
(d) A circular fountain has radius $4$ m, centred at origin.

Reveal solution

(a) LINEAR (constant rate), not in our four non-linear families. $V = 1200 - 20t$. ($y$-int $= 1200$ L initial volume.) (b) Parabola with vertex $(0, 8)$, $x$-ints $(-10, 0)$ and $(10, 0)$: $y = a x^2 + 8$ with $0 = 100a + 8 \Rightarrow a = -0.08$, so $y = -0.08 x^2 + 8$. $y$-int $= 8$ m = max arch height. (c) Exponential decay: $M = 80 \cdot \left(\tfrac{1}{2}\right)^{t/5}$. $y$-int $= 80$ g (starting mass). Asymptote $M = 0$: mass approaches zero. (d) Circle: $x^2 + y^2 = 16$. No $y$-intercept "meaning" — intercepts $(0, \pm 4)$ are just points on the fountain edge.

Quick Review

Projectile

Parabola: vertex = max

Inverse

Hyperbola: $xy = k$

Growth/decay

Exponential: $\times b$ each step

Circular path

Circle: distance $= r$

Reject

Negative time, neg people

Units

Always state them

Your Badges

0 of 6

First Steps

3-Day Streak

3 in a Row

Lesson Ace

Stretch Seeker

Daily Warrior

Mark lesson as complete

Tick when you've finished Learn, Practice and the Stretch. Earns +85 XP and +25 coins.

Unit overview · Maths subject page